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Pressure inside two soap bubbles are `1.01` and `1.02` atmospheres. Ratio between their volumes isA. 16B. 8C. 4D. 2

Answer» Correct Answer - B
Excess pressure in one bubble,
`p_(1)=1.01 - 1=0.01 =(4S)/(r_(1))` …(i)
Excess pressure in other bubble,
`p_(2)=1.02 - 1=0.02=(4S)/(r_(2))` ….(ii)
Dividing (i) by (ii)
`(0.01)/(0.02)=(r_(2))/(r_(1))` or `(r_(1))/(r_(2))=2`
`(V_(1))/(V_(2))=(r_(1)^(3))/(r_(2)^(3))=(2)^(3) =8`


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